Question

the magnitude of two vectors p bar ad q bar differ by 1. the magnitude of their resultant makes an angle of tan inverse (3/4) with p. the angle between p and q is

Answer 1

Answer on Question #44021 – Physics – Mechanics-Kinematics-Dynamics

the magnitude of two vectors $p$ bar and $q$ bar differ by 1. the magnitude of their resultant makes an angle of tan inverse (3/4) with $p$. the angle between $p$ and $q$ is

Solution:

$\alpha = \arctan \left(\frac{3}{4}\right)$ – angle between vector $p$ and resultant vector;

$\beta$ – angle between vector $p$ and vector $q$;

First vector:

$$p = (p_x, p_y), \text{ magnitude: } P = \sqrt{p_x^2 + p_y^2}$$

Second vector:

$$q = (q_x, q_y), \text{ magnitude: } Q = \sqrt{q_x^2 + q_y^2}$$

Let's make a substitution:

$$p_x q_x + p_y q_y = X$$

Difference between magnitude of two vectors:

$$P - Q = 1$$ $$Q = P - 1$$

Resultant vector:

$$\vec{r} = \vec{p} + \vec{q}$$ $$r = (p_x + q_x, p_y + q_y)$$

Scalar product of the first vector and resultant vector:

$$\vec{p} \cdot \vec{r} = p_x (p_x + q_x) + p_y (p_y + q_y) = |\vec{p}| \cdot |\vec{r}| \cdot \cos \alpha = P \cdot \sqrt{(p_x + q_x)^2 + (p_y + q_y)^2} \cdot \cos \alpha$$ $$p_x^2 + p_y^2 + p_x q_x + p_y q_y = P \cdot \sqrt{(p_x + q_x)^2 + (p_y + q_y)^2} \cdot \cos \alpha$$ $$P^2 + X = P \cdot \sqrt{(p_x + q_x)^2 + (p_y + q_y)^2} \cdot \cos \alpha = P \cdot \sqrt{p_x^2 + 2 p_x q_x + q_x^2 + p_y^2 + 2 p_y q_y + q_y^2} \cdot \cos \alpha$$ $$= P \cdot \sqrt{P^2 + Q^2 + 2X} \cdot \cos \alpha$$

Scalar product of the first vector and second vector:

$$\vec{p} \cdot \vec{q} = p_x q_x + p_y q_y = |\vec{p}| \cdot |\vec{q}| \cdot \cos \beta = P \cdot Q \cdot \cos \beta$$ $$\cos \beta = \frac{p_x q_x + p_y q_y}{P \cdot Q} = \frac{X}{P \cdot Q}$$

Thus, we have system with three equations:

$$\left\{ \begin{array}{c} Q = P - 1 \quad (1) \\ P^2 + X = P \cdot \sqrt{P^2 + Q^2 + 2X} \cdot \cos \alpha \\ \cos \beta = \frac{X}{P \cdot Q} \end{array} \right.$$

We have 3 equations and 4 unknown (P, Q, X and cos β), hence we can't find cos β – cosine of angle between vector p and vector q.

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